MCQs
Total Questions : 70
| Page 4 of 7 pages
Question 31. Find the heat flow rate through the composite wall as shown in figure. Assume one dimensional flow and take
k 1 = 150 W/m degree
k 2 = 30 W/m degree
k 3 = 65 W/m degree
k 4 = 50 W/m degree
AB = 3 cm, BC = 8 cm and CD = 5 cm.
The distance between middle horizontal line from the top is 3 cm and from the bottom is 7 cm
k 1 = 150 W/m degree
k 2 = 30 W/m degree
k 3 = 65 W/m degree
k 4 = 50 W/m degree
AB = 3 cm, BC = 8 cm and CD = 5 cm.
The distance between middle horizontal line from the top is 3 cm and from the bottom is 7 cm
Answer: Option B. -> 1273.88 W
Answer: (b).1273.88 W
Answer: (b).1273.88 W
Question 32. The oven of an electric store, of total outside surface area 2.9 m² dissipates electric energy at the rate of 600 W. The surrounding room air is at 20 degree Celsius and the surface coefficient of heat transfer between the room air and the surface of the oven is estimated to be 11.35 W/m² degree. Determine the average steady state temperature of the outside surface of the store
Answer: Option A. -> 38.22 degree Celsius
Answer: (a).38.22 degree Celsius
Answer: (a).38.22 degree Celsius
Question 33. A steel pipe of 20 mm inner diameter and 2 mm thickness is covered with 20 mm thick of fiber glass insulation (k = 0.05 W/m degree). If the inside and outside convective coefficients are 10 W/m² degree and 5 W/m² degree, calculate the overall heat transfer coefficient based on inside diameter of pipe. In the diagram, the diameter of small circle is 20 mm
Answer: Option B. -> 2.789 W/m² degree
Answer: (b).2.789 W/m² degree
Answer: (b).2.789 W/m² degree
Question 34. The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material?
Answer: Option B. -> 0.71 W/m k
Answer: (b).0.71 W/m k
Answer: (b).0.71 W/m k
Answer: Option A. -> d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)
Answer: (a).d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)
Answer: (a).d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)
Answer: Option B. -> Cylinder
Answer: (b).Cylinder
Answer: (b).Cylinder
Answer: Option B. -> Q = A k₀ (T₁³ – T₂³)/3 δ
Answer: (b).Q = A k₀ (T₁³ – T₂³)/3 δ
Answer: (b).Q = A k₀ (T₁³ – T₂³)/3 δ
Answer: Option A. -> Q = -k0 (1 + β t) A d t/d x
Answer: (a).Q = -k0 (1 + β t) A d t/d x
Answer: (a).Q = -k0 (1 + β t) A d t/d x
Question 39. The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation
K = (1.45 + 0.5 * 10¯⁵ t²) KJ/m hr deg
Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
K = (1.45 + 0.5 * 10¯⁵ t²) KJ/m hr deg
Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
Answer: Option C. -> 1745.8 kJ/m² hr
Answer: (c).1745.8 kJ/m² hr
Answer: (c).1745.8 kJ/m² hr
Answer: Option D. -> km = k0 [1 + β (t1 + t2)/2].
Answer: (d).km = k0 [1 + β (t1 + t2)/2].
Answer: (d).km = k0 [1 + β (t1 + t2)/2].